0.22 Entropy and the second law of thermodynamics  (Page 9/9)

Let’s observe the entropy changes for freezing water at -10 ºC, which we know to be a spontaneous process. First, let’s calculate the ∆S for the process H ₂ O( l ) → H ₂ O( s ) near 0 ºC. We can do this from a table of absolute entropies, such as in the previous Concept Development Study. At 0 ºC, S for H ₂ O( l ) = 63.3 J/mol·K and S for H ₂ O( s ) = 41.3 J/mol·K, so for the freezing of water, ∆S = -22.0 J/mol·K. As expected, this is less than zero.

We know from our previous observation that, since freezing releases heat into the surroundings, then freezing must increase the entropy of the surroundings. Let’s calculate this change in entropy, ∆S _surroundings . We will assume quite reasonably and fairly generally that the only effect on the entropy of the surroundings of the freezing taking place in the system is the transfer of heat from the system to the surroundings. There is no other exchange or interaction which happens. We might expect that releasing heat into the surroundings would change the temperature of the surroundings. However, the surroundings are typically huge (e.g. the entire room in which the process occurs) that the effect of the heat transfer into the surroundings does not make a measurable change in the surrounding temperature. The same is true when energy is absorbed from the surroundings into the system.

From our Concept Development Study on energy changes, we can calculate the heat transfer for a process occurring under constant pressure from the enthalpy change for the process, q = ΔHº. By conservation of energy, the heat flow out of or into the surroundings must be –ΔHº. From our previous observation, the entropy change resulting from this transfer of heat is

∆S° _surroundings = q/T = -∆H°/T

In this equation, ∆H is the enthalpy change for the system, which is why there is a negative sign in this equation. Notice that, if the process releases energy as is the case in freezing, ∆H°<0 so ∆S° _surroundings >0.

For H ₂ O( l ) → H ₂ O( s ), ∆Hº = ∆Hº _f (H ₂ O(s)) - ∆Hº _f (H ₂ O( l )). From the data tables, we can calculate that ∆Hº = -6.02 kJ/mol. As expected, this is less than zero. From this, we can calculate that at -10 ºC = 263 K, ∆S° _surroundings = 22.9 J/mol·K. Again, as expected, this is greater than zero.

We now come to the most important observation: the entropy increase of the surroundings is greater than the entropy decrease of the water that is freezing. This means that, in total, the entropy change for both system and surroundings is positive.

The conclusion of this observation is that, in analyzing whether a process is spontaneous or not, we must consider both the change in entropy of the system undergoing the process and the effect of the heat released or absorbed during the process on the entropy of the surroundings. In other words, we have to consider the entropy change of the universe:

∆S _universe = ∆S _system + ∆S _surroundings

We can say then that

∆S _universe = ∆S _system + ∆S _surroundings = ∆S _system - ∆H _system /T

The last part of this equation says that we can calculate the entropy change of the universe from calculations that involve only entropy and energy changes in the system. This sounds very doable! For every spontaneous process, ∆S _universe >0, so for every spontaneous process:

∆S _system - ∆H _system /T>0

In this inequality, the entropy change and the enthalpy change are both for the system, so the subscripts are now not needed and we can drop them to make the equation easier to read.

∆S - ∆H/T>0 any spontaneous process

The fact that the temperature appears in this equation is very interesting. Remember why it is there in the denominator: the effect on the entropy of the heat exchange with the surroundings is smaller when the temperature is larger.

This also suggests that we try our calculation with a different temperature. What if the temperature is 10 ºC, above the freezing point? In this case, we observe experimentally that the melting of solid ice is spontaneous. Let’s do the calculation of the inequality above for the process H ₂ O( l ) → H ₂ O( s ) at 10 ºC. As before, ∆Hº = -6.02 kJ/mol and ∆Sº = -22.0 J/mol·K, but now T = 283K. We get that

∆S° - ∆H°/T = -0.73 J/mol·K<0

We find a negative number, telling us that freezing water is not spontaneous at 10 ºC. More importantly, we know that the reverse process of melting ice is spontaneous at 10 ºC. This means that, if ∆S - ∆H/T<0, the reverse process is spontaneous.

Now we know what is so special about the temperature 0 ºC. Below that temperature, ∆S° - ∆H°/T>0 for the freezing of water, and above that temperature ∆S° - ∆H°/T<0 for the freezing of water, so ∆S° - ∆H°/T>0 for the melting of ice.

What happens at 0 ºC? It is easy to calculate that ∆S° - ∆H°/T = 0 at that temperature. Of course, at that temperature at standard pressure, water and ice are in equilibrium. This is a very important observation: when ∆S - ∆H/T = 0, the process is at equilibrium. We will develop this observation in much greater detail in the next Concept Development Study.

Review and discussion questions

1. Each possible sequence of the 52 cards in a deck is equally probable. However, when you shuffle a deck and then examine the sequence, the deck is never ordered. Explain why in terms of microstates, macrostates, and entropy.
2. Assess the validity of the statement, "In all spontaneous processes, the system moves toward a state of lowest energy." Correct any errors you identify.
3. Recalling the discussion of the freezing point of water, determine what must be true about ΔHº and ΔSº for a solid to have a freezing point much higher than that of water. Similarly, determine what must be true about ΔHº and ΔSº for a solid to have a freezing point much lower than that of water.
4. Predict the sign of the entropy for the reaction 2 H ₂ (g) + O ₂ (g) → 2 H ₂ O (g). Give an explanation, based on entropy and the Second Law, of why this reaction occurs spontaneously.
5. For the reaction H2(g) → 2 H(g), predict the sign of both ΔHº and ΔSº. Based on your knowledge of hydrogen, which of these two terms is probably dominant in determining whether the reaction is spontaneous. What would you predict would be the effect of lowering the temperature? Explain.